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3 years ago

The nearest star to Earth is Proxima Centauri, 4.3 light-years away.

Physics

1 answer:

PIT_PIT [208]3 years ago

7 0

Answer:

a) v = 2,152 10⁸ m / s b) t = 2.71 10⁸ s or t = 85.93 year

Explanation:

a) In this special relativity exercise we have that time is measured in the same ship, so it is the proper time,

v = d / t

Let's reduce the distance to the SI system

d = 4.3 l and (9.46 1015 m / 1ly) = 40.678 10¹⁵ m

t = 5.0 y (365 day / 1 y) (24 h / 1 day) (3600s / 1h) = 1.89 10⁸ s

Let's calculate

v = 40.678 10¹⁵ / 1.89 10⁸

v = 2,152 10⁸ m / s

b) The time seen from the ground for which the ship moves is given by

t = t₀ / √ (1- (v/c)²)

Let's calculate

t = 1.89 10⁸ / √ (1 - (2.152 / 2.998)²)

t = 1.89 10⁸ / 0.6962

t = 2.71 10⁸ s

Let's reduce this time to years

t = 2.71 10⁸ s (1h / 3600s) (1day / 24h) (1 and / 365 d)

t = 85.93 year

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3 years ago

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3 years ago

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3 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago